1. Field of the Invention
This disclosure relates to improved configurations for solid-state cooling, heating and power generation systems.
2. Description of the Related Art
Thermoelectric devices (TEs) utilize the properties of certain materials to develop a temperature gradient across the material in the presence of current flow. Conventional thermoelectric devices utilize P-type and N-type semiconductors as the thermoelectric material within the device. These are physically and electrically configured in such a manner that the desired function of heating or cooling is obtained.
The most common configuration used in thermoelectric devices today is illustrated in FIG. 1A. Generally, P-type and N-type thermoelectric elements 102 are arrayed in a rectangular assembly 100 between two substrates 104. A current, I, passes through both element types. The elements are connected in series via copper shunts 106 saddled to the ends of the elements 102. A DC voltage 108, when applied, creates a temperature gradient across the TE elements. TEs are commonly used to cool liquids, gases and solid objects.
Solid-state cooling, heating and power generation (SSCHP) systems have been in use since the 1960's for military and aerospace instrumentation, temperature control and power generation applications. Commercial usage has been limited because such systems have been too costly for the function performed, and have low power density so SSCHP systems are larger, more costly, less efficient and heavier than has been commercially acceptable.
Recent material improvements offer the promise of increased efficiency and power densities up to one hundred times those of present systems. However, Thermoelectric (TE) device usage has been limited by low efficiency, low power density and high cost.
It is well-known from TE design guides (Melcor Corporation “Thermoelectric Handbook” 1995 pp. 16-17) that in today's TE materials, the cooling power at peak efficiency produced by a module with ZT=0.9 is about 22% of the maximum cooling power. Thus, to achieve the highest possible efficiency, several TE modules are required compared to the number required for operation at maximum cooling. As a result, the cost of TE modules for efficient operation is significantly higher and the resulting systems are substantially larger.
It is known from the literature (for example, see Goldsmid, H. J. “Electronic Refrigeration” 1986, p. 9) that the maximum thermal cooling power can be written as;
                              q          COPT                =                                            I              OPT                        ⁢                          α              C                                -                                    1              2                        ⁢                          I              OPT              2                        ⁢            R                    -                      K            ⁢                                                  ⁢            Δ            ⁢                                                  ⁢            T                                              (        1        )            Where
qCOPT is the optimum cooling thermal power
IOPT is the optimum current
α is the Seebeck Coefficient
R is the system electrical resistance
K is the system thermal conductance
ΔT is the difference between the hot and cold side temperatures
TC is the cold side temperature
Further, from Goldsmid's,
                              I          OPT                =                                            α              R                        ⁢                          1                              (                                                                            ZT                      AVE                                        -                    1                                                  )                                              =                      α                          R              ⁡                              (                                  M                  -                  1                                )                                                                        (        2        )            WhereZ is the material thermoelectric figure of meritTAVE is the average of the hot and cold side temperaturesM=√{square root over (ZTAVE+1)}Substitution Equation (2) into (1) yields
                              q          OPT                =                              [                                                                                ZT                    C                                                        (                                          M                      -                      1                                        )                                                  ⁢                                  (                                                                                    Δ                        ⁢                                                                                                  ⁢                        T                                                                    T                        C                                                              -                                          1                                              2                        ⁢                                                  (                                                      M                            -                            1                                                    )                                                                                                      )                                            -                              Δ                ⁢                                                                  ⁢                T                                      ⁢                                                  ]                    ⁢          K                                    (        3        )            
The term on the right side of Equation (3) in brackets is independent of the size (or dimensions) of the TE system, and so the amount of cooling qOPT is only a function of material properties and K. For the geometry of FIG. 1, K can be written as;
                    K        =                              λ            ⁢                                                  ⁢                          A              C                                            L            C                                              (        4        )            Where λ is the average thermal conductivity of the N & P materials
AC is the area of the elements
L is the length of each element
Since α is an intrinsic material property, as long as the ratio Lc/Ac is fixed, the optimum thermal power qOPT− will be the same. For current equal to IOPT, the resistance is;
                              R          C                =                                            R              OC                        +                          R              PC                                =                                                                      ρ                  TE                                ⁢                                  L                  C                                                            A                C                                      +                          R              PC                                                          (        5        )            Where ρTE is the intrinsic average resistivity of the TE elementsROC is the TE material resistanceRPC is parasitic resistances
For the moment, assume RP is zero, then R is constant. IOPT is constant if LC/AC is fixed. Only if the ratio Lc/Ac changes, does K and hence, qCOPT and ROC and hence, IOPT changes.
Generally, it is advantageous to make a device smaller for the same cooling output. An important limitation in thermoelectric systems is that as, for example, the length LC is decreased for fixed AC, the ratio of the parasitic resistive losses to TE material losses, φC becomes relatively large.
                              ϕ          C                =                              R            PC                                R            OC                                              (        6        )            
This can be seen by referring to FIG. 1C, which depicts a typical TE couple. While several parasitic losses occur, one of the largest for a well-designed TE is that from shunt 106. The resistance of shunt 106 per TE element 102 is approximately,
                              R          PC                ≈                              (                                                            B                  C                                +                                  G                  C                                                                              W                  C                                ⁢                                  T                  C                                                      )                    ⁢                      P            SC                                              (        7        )            Where GC is the gap between the TE elements.BC is the TE element and shunt breadth.WC is the TE element and shunt width.TC is the shunt thickness.PSC is the shunt resistivity.
For the geometry of FIG. 1, the resistance for a TE element is
                              R          OC                =                                            P              TE                        ⁢                          L              C                                                          B              C                        ⁢                          W              C                                                          (        8        )            Where;Lc, is the TE element length.Thus, using Equations (7) and (8) in (6),
                              ϕ          C                ≈                                            B              C                        ⁡                          (                                                                    B                    C                                    +                                      G                    C                                                                                        T                    C                                    ⁢                                      L                    C                                                              )                                ⁢                      (                                          P                SC                                            P                TE                                      )                                              (        9        )            